Passed all Assessments. BUT Still confused on Simplifying Boolean Algebra

I passes all my assessments!!
Super excited/terrified to start the cohart.

However, I find that I still don’t understand a certain type of question regarding simplifying Boolean Algebra with DeMorgans Law.

I understand DeMorgans Law, but I think i’m getting hung up when there are multiple NOT symbols over a paranthesis, over another paranthesis …then again over another paranthesis.

A questions looks like this:
(for the sake of the assessments, i’ve changed the order of expressions, changed variables and what they represent. Also the NOT (’) symbols have been shuffled.)

If A is false, B is true, and Y is false. What is
((a and b’ and c) AND (a’ and b and c’))’ AND ((a and b and c) OR (a’ and b’ and c))

on the ((a and b’ and c) AND (a’ and b and c’))’ This is how I broke it down:

((ab’c)(a’bc’))’
((a’ + b + c’)(a+b’c))

or is there a break down of paranthesis like this?

((ab’c)(a’b’c’))’
(a’+b+c’+a+b+c)

or …am I not even close?

any help?

thanks,

1 Like

Ultimately, what is the question asking? For a simplified expression, or true/false?

it’s a true/false.
However, i’m unsure the method on how to simplify this problem with Demorgans law expressions.

I would say make sure you have a basic understanding of Demorgan’s Law a’b’= (a+b)’. What I find that helps is to say the expression using “and” “or” and “not” (not a AND not b equals not a OR b) so neither. obviously sub 1 for true and 0 for false and plug them in to find the solution. Start with your innermost equation/expression and work your way outward, so if the solution inside on set of parenthesis is (1) and there is a final ’ on the outside of that original equation, you would apply that ’ to the 1, which would mean NOT 1, which is 0. Does that make sense?

Congrats on passing all of your assesments!!